解:∵abc不等于0 且a+b+c不等于0
c/(a+b)=a/(b+c)=b/(a+c)
由等比性质,得
(a+b+c)/(2a+2b+2c)=k
∴k=(a+b+c)/(2a+2b+2c)
=(a+b+c)/[2(a+b+c)]
=½
∴k-1=½-1=-½
则反比例函数y=(k-1)/x的图像经过【第二、四】象限.
:∵abc不等于0 且a+b+c不等于0
c/(a+b)=a/(b+c)=b/(a+c)
由等比性质,得
(a+b+c)/(2a+2b+2c)=k
∴k=(a+b+c)/(2a+2b+2c)
=(a+b+c)/[2(a+b+c)]
=½
∴k-1=½-1=-½